Problem: Solve for $x$ : $5x^2 + 15x - 140 = 0$
Dividing both sides by $5$ gives: $ x^2 + {3}x {-28} = 0 $ The coefficient on the $x$ term is $3$ and the constant term is $-28$ , so we need to find two numbers that add up to $3$ and multiply to $-28$ The two numbers $7$ and $-4$ satisfy both conditions: $ {7} + {-4} = {3} $ $ {7} \times {-4} = {-28} $ $(x + {7}) (x {-4}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 7) (x -4) = 0$ $x + 7 = 0$ or $x - 4 = 0$ Thus, $x = -7$ and $x = 4$ are the solutions.